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\begin{document}

\section{Problem Sheet 1}

\subsection{Capacitor}

\subsubsection{a)}

Let the capacitor have area $A$, with plate separation $d$. The volume of the inside of the capacitor is thus $V = Ad$.

Assume that $\E$ field is zero outside the capacitor.

The momentum density stored within the fields is $$\vec p(\r) = \frac{1}{c} \E(\r) \times \B(\r)$$ hence the total momentum is the density integrated over the region where $\vec E \times \vec B$ is non-zero (i.e. the inside of the capacitor) is simply $$\vec P = \int \d V' \vec p(\r) = \frac{Ad}{c} \E \times \B = \frac{Ad}{c} EB \vec e_z \times \vec e_x = \frac{Ad}{c} EB \vec e_y$$

The force on a conductor $l$ is $$\vec F = \frac{1}{c} \int_l \d r' \J(\r') \times \B(\r')$$

The current $\J(\r)$ at any point in the conductor is $$\J = \diff{Q}{t} \vec e_z$$ where $Q$ is the charge on the plates of the capacitor. Hence the force reduces to $$\vec F = \frac{1}{c} \int_l \d r' \diff{Q}{t} \vec e_z \times B \vec e_x = \diff{Q}{t} \frac{Bd}{c} \vec e_y$$

The total impulse is the force integrated over the time taken for the capacitor to discharge $$\vec P = \int \d t \diff{Q}{t} \frac{Bd}{c} \vec e_y = \diff{}{t}\int \d t \frac{Q B d}{c} \vec e_y = \frac{QBd}{c} \vec e_y$$

Now compare with the previous expression for $\vec P$. Insert the definition of capacitance $Q = CV$ into the above expression, remembering that for a parallel plate capacitor $C = A / d$, $V = E d$, to obtain $C = A E$. Thus $$\vec P = \frac{AdEB}{c} \vec e_y$$ as required.

\subsubsection{b)}

The wire is now absent, hence the force must be exerted directly upon the charged plates themselves. The changing magnetic field will induce a 	

\section{Boundary Impedance}

\section{Monopoles}

Angular momentum density is

$$j_{em}(\r) = \r \times \vec p(\r) = \frac{1}{c} \r \times \left( \E(\r) \times \B(\r) \right)$$

hence the total angular momentum (measured about the origin) is this quantity, integrated over the extent of the fields (i.e. all space)

$$J_{em} = \frac{1}{c}  \int \d V'  \r' \times \left( \E(\r') \times \B(\r') \right)$$

Now insert explicit expressions for $\E = q (\r - \vec a) / (4 \pi | \r - \vec a |^3)$ and $\B(\r) = g \r / (4 \pi r^3)$

$$J_{em} = \frac{qg}{16 \pi^2 c}  \int \d V'  \r' \times \left( \frac{ (\r' - \vec a) \times \r' }{ |\r' - \vec a|^3 r^3 } \right)$$

\subsection{Dynamic Fields from a Straight Wire}

The current is spatially confined to the wire, which we take to be oriented in the $z$ direction, hence $$\J(\r) = I_0 \delta(x-x') \delta(y-y')  \hat{\vec z}$$ now if we assume that the current propagates instantaneously through the wire (although this is unphysical), then the time dependence is described by a step function centred at $t=0$ $$\J(\r,t) = I_0 \delta(x-x')\delta(y-y') \theta(t) \hat{\vec z}.$$ The resulting vector potential is given by the integral $$\A(\r,t) = \frac{1}{4 \pi} \int \d V' \frac{\J(\r,\tau)}{|\r - \r'|}$$ which is spatially restricted to an integral along the wire, so that $$|\r - \r'| = \sqrt{ (z - z')^2 + x^2 + y^2 }$$ and the integral becomes $$\A(\r,t) = \frac{1}{4 \pi} \hat{\vec z} \int_{-\infty}^{\infty} \d z' \frac{I_0 \theta(t - \sqrt{ (z - z')^2 + x^2 + y^2 }/c)}{\sqrt{ (z - z')^2 + x^2 + y^2 }} $$ so the field `switches on' when the argument of the theta function becomes greater than zero, i.e. when $$t -  \sqrt{ (z - z')^2 + x^2 + y^2 }/c = 0 \Rightarrow z' = z -\sqrt{(ct)^2 - x^2 - y^2}$$ and so the limits of integration are modified according to $$\A(\r,t) = \frac{1}{4 \pi} \hat{\vec z} \int_{z -\sqrt{(ct)^2 - x^2 - y^2}}^{\infty} \d z' \frac{I_0}{\sqrt{ (z - z')^2 + x^2 + y^2 }} $$

\subsection{Quasi-Statics}

a)

Starting with charge conservation $$\pdiff{\rho}{t} = - \div \J,$$ now if $\J = \J(\r)$, then the right-hand side is some function of $\r$ only, which we identify as $\dot \rho(\r)$ $$\pdiff{\rho}{t} = \dot \rho(\r)$$ so integrating between $t$ and $\tau$ $$\int_t^\tau \d t' \pdiff{\rho}{t'} = \int_t^\tau \d t' \dot \rho(\r)$$ $$\Rightarrow \rho(\r,\tau) - \rho(\r,t) = (\tau - t) \dot \rho(\r)$$ and rearranging, we obtain $$\Rightarrow \rho(\r,\tau) = \rho(\r,t) + (\tau - t) \dot \rho(\r)$$ as required.

Starting from the Jefimenko equations for electric field $$\E = \frac{1}{4 \pi} \int \d V' \left\{ \left( \frac{1}{c} \pdiff{\rho}{t} + \frac{\rho}{|\r - \r'|} \right) \frac{\r - \r'}{|\r - \r'|^2} - \frac{1}{c^2} \pdiff{\J}{t} \frac{1}{|\r - \r'|} \right\} $$ and magnetic field $$\B = \frac{1}{4 \pi c} \int \d V' \left\{ \left( \frac{1}{c} \pdiff{\J}{t} + \frac{\J}{|\r - \r'|} \right) \times \frac{\r - \r'}{|\r - \r'|^2} \right\}, $$ recall that these must depend on the configuration of the sources as they were at the \emph{retarded} time $$\tau = t - |\r - \r'|/c$$ and so substituting the result found above $$\rho(\r,\tau) = \rho(\r,t) + (\tau - t) \dot \rho(\r) = \rho(\r,t) - |\r - \r'| \dot \rho(\r) / c$$ and the corollaries $\pdiff{\rho}{t} = \dot \rho(\r)$ and $\J(\r,\tau) = \J(\r)$, we obtain for the electric field \begin{eqnarray*}
\E(\r,t) &=& \frac{1}{4 \pi} \int \d V' \left\{ \left( \frac{1}{c} \dot \rho(\r') + \frac{ [ \rho(\r',t) - |\r - \r'| \dot \rho(\r') / c] }{|\r - \r'|} \right) \frac{\r - \r'}{|\r - \r'|^2} - \frac{1}{c^2} \pdiff{}{t} \J(\r') \frac{1}{|\r - \r'|} \right\} \\
&=& \frac{1}{4 \pi} \int \d V' \rho(\r',t) \frac{(\r - \r')}{|\r - \r'|^3}
\end{eqnarray*} by cancellation of the terms in parentheses, and since the time derivative of $\J(\r)$ vanishes trivially. Similarly, we obtain for the magnetic field \begin{eqnarray*}
\B &=& \frac{1}{4 \pi c} \int \d V' \left\{ \left( \frac{1}{c} \pdiff{}{t} \J(\r') + \frac{\J(\r')}{|\r - \r'|} \right) \times \frac{\r - \r'}{|\r - \r'|^2} \right\} \\
&=& \frac{1}{4 \pi c} \int \d V' \J(r') \times \frac{ (\r - \r') }{ |\r - \r'|^3}
\end{eqnarray*} i.e. we recover Coulomb's Law and the Biot-Savart Law from statics.

We have thus shown that in magnetostatics ELECTROSTATICS STILL HOLDS??

b)

Now consider $\J$ is no longer a constant, but varying linearly with time, i.e. $$\J(\r,\tau) = \J(\r,\tau) + (\tau - t) \dot \J(\r)$$ $$\Rightarrow \J(\r,\tau) = \J(\r,\tau) - |\r-\r'| \dot \J(\r) / c,$$ then $$\pdiff{}{t}\J(\r,\tau) = \pdiff{}{t} \J(\r,t),$$ i.e. the $\tau$ dependence is removed, hence upon substituting $\J$ into Jefimenko's equation for the magnetic field we obtain \begin{eqnarray*}
\B &=& \frac{1}{4 \pi c} \int \d V' \left\{ \left( \frac{1}{c} \pdiff{}{t}\J(\r',t) + \frac{ [ \J(\r',t) - |\r - \r'| \dot \J(\r') / c ] }{ |\r - \r'| } \right) \times \frac{ \r - \r' }{ |\r - \r'|^2} \right\} \\
&=& \frac{1}{4 \pi c} \int \d V' \J(\r',t) \times \frac{ (\r - \r') }{ |\r - \r'|^3}
\end{eqnarray*} where we have identified $$\pdiff{}{t} \J(\r,t) = \dot \J (\r),$$ thus we have shown that the Biot-Savart law still holds for uniformly-accelerated charges.


\section{Problem Sheet 2}

\subsection{Gauge Transformations}

$\E = \grad \phi$, $\phi = 0$ implies that $\E = \grad 0 = 0$.

\begin{eqnarray*}
\B &=& \curl \vec A = -\frac{q}{4 \pi} ct \curl \frac{ \r }{ r } \\
&=& -\frac{q}{4 \pi} ct \left( \frac{1}{r} \curl \r + \r \times \grad (1/r) \right) \\
&=& -\frac{q}{4 \pi} ct \frac{1}{r^2} \r \times \r = 0
\end{eqnarray*}

(ii)

Gauge transformations:

$$\vec A \rightarrow \vec A + \grad \chi$$

$$\phi \rightarrow \phi + \frac{1}{c} \pdiff{\chi}{t}$$

So for given $\phi, \vec A, \chi$:

$$\grad \chi = \frac{q}{4 \pi} ct \frac{1}{r^3} \r$$

$$\vec A' = \frac{q}{4 \pi} ct \left( \frac{1}{r^3} \r - \frac{1}{r^3} \r \right) = 0$$

$$\phi' = - \frac{q}{4 \pi} \frac{1}{r}$$

The magnetic field $\B$ is clearly still zero. The electric field $E$ can be checked 

(iii)

Lorentz gauge requires that 

$$\div \vec A + \frac{1}{c} \pdiff{}{t} \phi = 0$$

and $\vec A$ and $\phi$ are \emph{already} in Lorentz gauge. Substituting the transformed potentials $\vec A' = \vec A + \grad \chi$ and $\phi' = \phi + \frac{1}{c} \pdiff{\chi}{t}$ into the Lorentz condition gives

$$\div \left(\vec A + \grad \chi \right) + \frac{1}{c} \pdiff{}{t} \left( \phi + \frac{1}{c} \pdiff{\chi}{t} \right) = 0$$
$$\div \vec A + \div ( \grad \chi ) + \frac{1}{c} \pdiff{\chi}{t} + \frac{1}{c^2} \pdiffsq{\chi}{t} = 0$$

and since the terms involving the original potentials $\vec A$ and $\phi$ sum to zero, we are left with the condition

$$\laplacian \chi + \frac{1}{c^2} \pdiffsq{\chi}{t} = 0$$

(since $\div ( \grad f ) = \laplacian f$ )i.e. a wave equation for $\chi$. Hence any periodic $\chi$ satisfying the wave equation will do. 

\subsection{Coulomb Gauge}

Maxwell's equations (in Lorentz-Heaviside):

$$\div \B = 0$$
$$\curl \B = \frac{1}{c} \pdiff{\E}{t} + \frac{1}{c} \J$$
$$\div \E = \rho$$
$$\curl \E = - \frac{1}{c} \pdiff{\B}{t}$$

Rewrite the curl equation for $\E$

$$\curl \left( \E +  \frac{1}{c} \pdiff{}{t} \A \right) = 0$$

and since the curl of the terms in brackets vanishes, this implies that they can be rewritten as the gradient of a scalar $\phi$

$$\E = - \grad \phi -  \frac{1}{c} \pdiff{}{t} \A$$

Now, using existence of vector potential $\B = \curl \A$, and the above expression for $\E$, rewrite the inhomogeneous magnetic equation

$$\curl ( \curl \A ) = \frac{1}{c} \pdiff{}{t} \left( - \grad \phi -  \frac{1}{c} \pdiff{}{t} \A \right) + \frac{1}{c} \J$$

Apply the expansion for the left-hand side, and write $\J = \grad \dot \phi + \J_2$

$$\grad ( \cancel{ \div \A } ) - \delsq \A = - \frac{1}{c^2} \pdiffsq{\A}{t} + \frac{1}{c} \J_2 $$

$$ \delsq \A - \frac{1}{c^2} \pdiffsq{\A}{t} = - \frac{1}{c} \J_2 $$

where $\div \A = 0$ is the Coulomb condition. This is an inhomogeneous wave equation for $\A$, which depends solely on $J_2$.

Using the inhomogeneous equation for $\E$

$$ \div \left( - \grad \phi -  \frac{1}{c} \pdiff{}{t} \A \right) = - \rho $$

$$\Rightarrow \delsq \phi + \frac{1}{c} \pdiff{}{t} \cancel{ \div \A } = \rho$$

$$\Rightarrow \delsq \phi = \rho$$

which is Poisson's equation for $\phi$

Since $\J_1 = \grad \dot \phi$ is the gradient of a scalar, its curl must vanish

$$\curl \J_1 = \curl \grad \dot \phi = 0$$

Proof:

Let $\vec f = \grad \psi$

$$\left[ \curl \vec f \right]_i = \epsilon_{ijk} \partial_j \partial_k \psi = - \epsilon_{ikj} \partial_j \partial_k \psi $$

where we have permuted the indices of the antisymmetric tensor. But partial derivatives commute, i.e. $\partial_j \partial_k \equiv \partial_k \partial_j$ hence we can switch the order of derivatives, and then relabel the dummy indices $k \leftrightarrow j$ to obtain

$$ \epsilon_{ijk} \partial_j \partial_k \psi = - \epsilon_{ijk} \partial_j \partial_k \psi $$

and so $\vec f$ must vanish.

Taking the divergence of the wave equation for $\A$

$$\div \delsq \A - \frac{1}{c^2} \pdiffsq{}{t} \cancel{ \div \A } = - \frac{1}{c} \div \J_2$$

the Laplacian of $\A$ consists of two terms $\curl (\curl \A) - \grad ( \div \A )$. The divergence term is zero by the Coulomb condition, and the divergence of a curl always vanishes, to give

$$\div \J_2 = 0$$

Proof:

Let $\vec f = \curl \vec g$

$$\div \vec f  =  \partial_i \epsilon_{ijk} \partial_j g_k =  - \partial_i \epsilon_{jik} \partial_j g_k $$

and similarly to above, by switching order of differentiation and relabelling dummy indices $i, j$

$$ \partial_i \epsilon_{ijk} \partial_j g_k =  - \partial_i \epsilon_{ijk} \partial_j g_k $$

and so $\vec f$ must vanish.

Using the Poisson equation for $\phi$

$$\delsq \phi = \rho$$

Realise that current conservation 

2.4

A mass falling under gravity has a trajectory given by

$$\vec z(t) = \hat{\vec z} \left( z_0 - \frac{1}{2} g t^2 \right)$$

Taking the origin to be immediately below the falling particle, then considering the charge and its image to be a dipole, the dipole moment of the configuration is

$$\vec p = q \vec z(t)$$

Larmor's formula gives the total power radiated by a dipole

$$P = \frac{\ddot p^2}{6 \pi c^3}$$

inserting the relevant dipole moment $\ddot{ \vec p} = - qgt \hat{\vec z}$

$$P = \frac{q^2 g^2}{6 \pi c^3}$$

2.5

Taking the origin at the equilibrium point of the particle, the trajectory of the oscillating particle is

$$\vec z(t) = - \hat{\vec z} d \cos( \omega t )$$

where $\omega = \sqrt{ k / m }$. As before, we consider the particle as a dipole with moment

$$\vec p = q \vec z(t)$$

The Poynting vector for dipole radiation is

$$\vec S = \frac{1}{16 \pi^2 c^3 r^2} \ddot{p}^2 \sin^2\theta \hat{\vec r}$$

again, inserting the relevant dipole moment $\ddot{\vec p} = \hat{\vec z} d \omega^2 \cos (\omega t)$, the power radiated is

$$\vec S = \frac{1}{16 \pi^2 c^3 r^2} d \omega^4 \cos^2 (\omega t)  \sin^2\theta \hat{\vec r}$$

a)

In our chosen geometry, $r^2 = R^2 + h^2$ (i.e. a right triangle), and so $r = h \cos \theta$, and the radiation impinging on the floor at angle $\theta$ is

$$\vec S = \frac{1}{16 \pi^2 c^3 h^2} d \omega^4 \cos^2 (\omega t)  \tan^2\theta \hat{\vec r}$$

$\tan^2 \theta$ is maximised for $\theta = \pi / 2$, hence the intensity of radiation is maximised where $\theta = \pi / 2 \Rightarrow R = h$. This is as expected, recalling the toroidal shape of the surface of equal energy density around an oscillating dipole.

b)

For the total radiation impinging on a floor of infinite extent, we have the Larmor formula, with an additional factor of $1/2$ (since we are integrating over a half-sphere, rather than a full sphere)

$$\langle P \rangle = \frac{w^4 p_0^2}{12 \pi c^3}$$

where $w = \sqrt{k/m}$ is the angular frequency of the oscillation, and $p_0 = d$ is the amplitude of the oscillation, hence in our case

$$\langle P \rangle = \frac{k^2 d^2}{12 \pi m^2 c^3}$$

c)

A mass undergoing SHM with maximum amplitude $d$ has total energy

$$E_0 = \frac{1}{2} m \omega^2 d^2$$

and with energy lost at a rate $\langle P \rangle$, after a time $\tau$

$$E_0 \left(1 - \frac{d}{e} \right) = \tau \langle P \rangle$$

Using the Larmor formula from (b) for $\langle P \rangle$, and solving for $\tau$

$$\tau = $$

\subsection{Rotating Dipole}

The approximations derived for the electric and magnetic fields from dipole radiation are linear in $\vec p$, however the Poynting vector $$\vec S = c \E \times \B = \frac{1}{16 \pi^2 c^3 r^2} \left[ \hat \r \times \ddot \vec p \right]^2 \hat \r$$ is quadratic in $\vec p$, hence we cannot simply use the formula $$\vec S = \frac{1}{16 \pi c^3 r^2} | \ddot \vec p |^2 \sin^2 \theta \hat \r,$$ as this is only valid for dipoles whose orientation does not change in time. For the given rotating dipole moment $$\vec p = p_0 \left[ \vec{ \hat x} \sin(\omega t) + \vec{ \hat y} \cos(\omega t) \right],$$ the acceleration of the dipole is $$\ddot \vec p = - p_0 w^2 \left[ \vec{ \hat x} \sin(\omega t) + \vec{ \hat y} \cos(\omega t) \right].$$ Now to evaluate $\hat \r \times \ddot \vec p$, consider $\hat \r = \frac{1}{r} \left[ x \hat{\vec x} + y \hat{\vec y} + z \hat{\vec z} \right]$ \begin{eqnarray*}
\hat \r \times \ddot \vec p &=&  - p_0 w^2 \frac{1}{r} \left[ x \hat{\vec x} + y \hat{\vec y} + z \hat{\vec z} \right] \times \left[ \vec{ \hat x} \sin(\omega t) + \vec{ \hat y} \cos(\omega t) \right] \\
&=& - \frac{p_0 \omega^2}{r} \left|\begin{array}{ccc} \hat{\vec x} & \hat{\vec y} & \hat{\vec z} \\x & y & z \\ \sin(\omega t) & \cos(\omega t) & 0 \end{array}\right| \\
&=& - \frac{p_0 \omega^2}{r} \left[ - \hat{\vec x} z \cos(\omega t) + \hat{\vec y} z \sin(\omega t) + \hat{\vec z} \left( x \cos(\omega t) - y \sin(\omega t) \right) \right].
\end{eqnarray*}

Squaring the above causes several useful cancellations \begin{eqnarray*}
\left[ \hat \r \times \ddot \vec p \right]^2 &=&  \frac{p^2_0 w^4}{r^2} \left[ z^2 \cos^2(\omega t) + z^2 \sin(\omega t) + \left( x \cos(\omega t) - y \sin(\omega t) \right)^2 \right] \\
&=& \frac{p^2_0 w^4}{r^2} \left[ z^2 - 2 xy \cos(\omega t) \sin(\omega t) + x^2 \cos^2 (\omega t) + y^2 \sin^2(\omega t) \right]
\end{eqnarray*} If we now time average the above, we pick up factors of $1/2$ from the squared terms, and the cross term involving $xy$ vanishes $$\langle \left[ \hat \r \times \ddot \vec p \right]^2 \rangle = \frac{p^2_0 \omega^4}{2 r^2} \left[ x^2 + y^2 + 2z^2 \right]$$ hence the time-averaged Poynting vector is $$\langle \vec S \rangle = \frac{p^2_0 \omega^4}{32 \pi^2 c^3 r^4} \left[ x^2 + y^2 + 2z^2 \right] \hat \r.$$

To find the total power radiated, it is easiest to perform the integral in spherical polar coordinates, hence we change variables according to $$x = r \sin\theta \cos \phi, \, y = r \sin \theta \sin \phi, \, z = r \cos \theta$$ which yields the integral \begin{eqnarray*}
\int_S \langle \vec S \rangle \cdot \d \A &=& \frac{p^2_0 \omega^4}{32 \pi^2 c^3 r^4} \hat \r \int_0^\pi \int_0^{2 \pi} \left[ r^2 \sin^2 \theta \cos^2 \phi + r^2 \sin^2 \theta \sin^2 \phi + 2 r^2 \cos^2 \theta \right] r^2 \sin \theta \d \theta \d \phi \\
 &=& \frac{p^2_0 \omega^4}{32 \pi^2 c^3} \hat \r \int_0^\pi \int_0^{2 \pi} \left[  \sin^2 \theta  + 2 \cos^2 \theta \right] \sin \theta \d \theta \d \phi \\
 &=& \frac{p^2_0 \omega^4}{32 \pi^2 c^3} \hat \r \int_0^\pi \int_0^{2 \pi} \left[  \sin \theta  + \sin \theta \cos^2 \theta \right] \d \theta \d \phi \\
\end{eqnarray*} now using the reduction formula $$- \frac{1}{n+1} \diff{}{\theta} ( \cos^{n+1} \theta ) = \sin \theta \cos^n \theta$$ the integral becomes \begin{eqnarray*}
\int_S \langle \vec S \rangle \cdot \d \A &=& \frac{p^2_0 \omega^4}{16 \pi c^3} \hat \r \int_0^\pi \left[  \sin \theta  - \frac{1}{3} \diff{}{\theta} ( \cos^3 \theta ) \right] \d \theta \\
 &=& \frac{p^2_0 \omega^4}{16 \pi c^3} \hat \r \left\{ \left[ - \cos \theta \right]_0^\pi  - \frac{1}{3} \left[ \cos^3 \theta \right]_0^\pi \right\} \\
  &=& \frac{p^2_0 \omega^4}{16 \pi c^3} \hat \r \left\{ \left[ 2 \right] - \frac{1}{3} \left[ - 2 \right] \right\}  = \frac{p^2_0 \omega^4}{3 \pi c^3} \hat \r 
\end{eqnarray*} i.e. the power radiated by a rotating dipole is twice the power radiated by an ordinary oscillating dipole.

Now for a hydrogen atom, time taken to radiate all rotational energy should be of order $10^-{10}$ seconds. Take care when converting to SI units.

\section{Relativistic Dynamics}

\subsection{Allowed Processes}

\textbf{Show that it is impossible for a free electron to emit a single photon within relativistic kinematics.}

Let $p_\mu$ be the initial electron 4-momentum, $q_\mu$ the final electron 4-momentum, and $r_\mu$ the photon 4-momentum, then $$p_\mu = q_\mu + r_\mu$$ $$\Rightarrow p_\mu p^\mu = q_\mu q^\mu + r_\mu r^\mu + 2 q_\mu r^\mu$$ by conservation of 4-momentum. Now by the definition of 4-momentum $$p_\mu = (\gamma m c, \gamma m \vec u),$$ then for \emph{any} massive particle $$p_\mu p^\mu = \gamma^2 m^2 c^2 - \gamma^2 m^2 \vec u^2 = \gamma^2 m^2 c^2 \left( 1 - \frac{\vec u^2}{c^2} \right) = m^2 c^2,$$ and since the mass $m$ is an intrinsic property of the electron, this is also true for the electron after it emits the photon, i.e. $$p_\mu p^\mu = q_\mu q^\mu = m^2 c^2.$$ For a massless particle $E^2 = \vec p^2 c^2$, hence the 4-momentum of a massless particle is $$r_\mu = \left( \frac{| \vec r | c}{c}, \vec r \right)= \left( | \r |, \r \right) \Rightarrow r_\mu r^\mu = \r^2 - \r^2 = 0$$ i.e. the 4-momentum of a massless particle is always null.

Substituting these into the above relationship, we obtain $$m^2 c^2 = m^2 c^2 + 0 + 2 q_\mu r^\mu \Rightarrow q_\mu r^\mu = 0.$$ Now since $q_\mu q^\mu = m^2 c^2$, $q_\mu$ must be a \emph{time-like} vector, hence in some basis it is proportional to $(1,0,0,0)$. The Lorentz scalar $$0 = q_\mu r^\mu \propto (1,0,0,0) \left(\begin{array}{c}r_0 \\r_1 \\r_2 \\r_3\end{array}\right) = r_0 \Rightarrow r_0 = 0$$ i.e. the zeroth component $r_0 = | \r | = 0$, and since $r_\mu$ is a null 4-vector, this is only possible if all components of $r_\mu = 0$. Physically, this is interpreted to mean that the photo represented by $r_\mu$ cannot exist.

\textbf{Show that if the electron omits more than one photon, these photons cannot all be parallel.}

As established previously, for a photon $$p_\mu = \left( |\vec p|, \vec p \right) \Rightarrow p_\mu p^\mu = | \vec p |^2 - \vec p^2 = 0.$$ The scalar product of two photons $p_\mu$ and $q_\mu$ is thus $$p_\mu q^\mu = | \vec p | | \vec q | - \vec p \cdot \vec q,$$ but if $p_\mu \propto q_\mu$ (i.e. if $p_\mu$ and $q_\mu$ are parallel) then $$\vec p \cdot \vec q = | \vec p | | \vec q | \Rightarrow p_\mu q^\mu = 0$$ i.e. their scalar product must vanish. Now consider a decay process in which an electron $p_\mu$ decays into another electron $q_\mu$ and emits \emph{several} photons, labelled by ${}^{(i)}r_\mu$; $$p_\mu = q_\mu + \sum_i {}^{(i)} r_\mu$$ by conservation of 4-momentum. To show that these photons cannot all be parallel, consider the following manipulation \begin{eqnarray*}
\left( p_\mu - q_\mu \right)^2 &=& \left( \sum_i {}^{(i)} r_\mu \right)^2 \\
\Rightarrow \left( p_\mu - q_\mu \right)^2 &=& \sum_{i \neq j} C_{ij} {}^{(i)}r_\mu {}^{(j)}r^\mu
\end{eqnarray*} where $C_{ij}$ are some (positive) constants of proportionality. Clearly individual photon momenta are self-orthogonal, and so terms in the expansion which are proportional to ${}^{(i)}r_\mu {}^{(i)}r^\mu = 0$ must vanish, leaving only those terms ${}^{(i)}r_\mu {}^{(j)}r^\mu, \; i \neq j$. Now if \emph{all} of the photons emitted are parallel, then all terms involving ${}^{(i)}r_\mu {}^{(j)}r^\mu = 0$ vanish also, and the right hand side will vanish identically, yielding $$( p_\mu - q_\mu )^2 = 0$$ i.e. $p_\mu - q_\mu$ is a null vector, but $p_\mu$, $q_\mu$ are both \emph{time-like} vectors, and no linear combination of time-like vectors can produce a null vector; hence contradiction, ${}^{(i)}r_\mu {}^{(j)}r^\mu \neq 0$ for some $i \neq j$, i.e. the photons cannot all be parallel.

\textbf{Show similarly that an electron and a positron cannot annihilate to produce a single real photon, or in fact any single real particle.}

Let $p_\mu$ be the electron, $q_\mu$ the positron, and $r_\mu$ the resulting particle, then $$p_\mu + q_\mu = r_\mu.$$ Rearrange and square the above to obtain \begin{eqnarray*}
p_\mu &=& r_\mu - q_\mu \\
\Rightarrow m^2 c^2 &=& r_\mu r^\mu + m^2 c^2 - 2 q_\mu r^\mu \\
\Rightarrow r_\mu r^\mu &=& 2 q_\mu r^\mu
\end{eqnarray*} now if $r_\mu$ is a photon, then $r_\mu r^\mu = 0$ and we have $$q_\mu r^\mu = 0,$$ and as previously shown, since $q_\mu$ is time-like, $r_\mu$ must be identically zero. For a general real particle, then $r_\mu$ is time-like and $r_\mu r^\mu > 0$, $q_\mu$ is also time-like

\subsection{Collisions}

Mulitply out each of the invariants and collect terms \begin{eqnarray*}
s + t + u &=& p_1^2 + p_2^2 + 2 p_1 p_2 + p_1^2 + p_3^2 - 2 p_1 p_3 + p_1^2 + p_4^2 - 2 p_1 p_4 \\
&=& p_1^2 + p_2^2 + p_3^2 + p_4^2  + 2 p_1^2 + 2p_1p_2 - 2p_1 p_3 - 2p_1 p_4 \\
&=& m_1^2 c^2 + m_2^2 c^2 + m_3^2 c^2 +m_4^2 c^2 + 2 p_1 ( p_1 + p_2 - p_3 - p_4 )
\end{eqnarray*} but by conservation of momentum $$p_1 + p_2 = p_3 + p_4 \Rightarrow p_1 + p_2 - p_2 - p_4 = 0$$ hence the terms in parentheses vanish and we are left with $$s + t + u = m_1^2 c^2 + m_2^2 c^2 + m_3^2 c^2 +m_4^2 c^2$$ as required.

The 4-momentum of the incoming particle is $$p_1 = \left( \frac{E_1}{c}, \vec p \right)$$ (where the subscript is a label, rather than an index). The stationary particle had 4-momentum $$p_2 = \left( \frac{E_2}{c}, 0 \right),$$ but $E_2 = m_2 c^2$, hence $$p_2 = \left( m_2 c, 0 \right).$$ The invariant $s$ is then $$s = (p_1 + p_2)^2 = p_1^2 + p_2^2 + 2 p_1 p_2 = m_1^2 c^2 + m_2^2 c^2 + 2 E_1 m_2,$$ but the particles are identical ($m_1 = m_2$) and so $$s = m^2 c^2 + m^2 c^2 + 2Em = 2m (E + mc^2)$$ as required. In the CoM frame, where both particles have the same energy $E_1 = E_2$, but opposing 3-momentum $\vec p_1 = - \vec p_2$, we have (paying attention to the sign of $p_1 p_2$) \begin{eqnarray*}
s &=& p_1^2 + p_2^2 + 2 p_1 p_2 \\
&=& m^2 c^2 + m^2 c^2 + 2 \left( \frac{E}{c}, \vec p \right) \left( \frac{E}{c}, - \vec p \right) \\
&=& 2 \left( m^2 c^2 + \frac{E^2}{c^2} + \vec p^2 \right)
\end{eqnarray*} and upon applying the relativistic energy-momentum relationship $E^2 = \vec p^2 c^2 + m^2 c^4$ we obtain $$s = 4 \frac{E^2}{c^2}$$ as required. 

\subsection{Scattering Angles}

\textbf{Show that when two identical particles of mass $m$ collide elastically in their centre-of-mass frame, they scatter through an angle $\phi$, with the same speed $v$ after the collision as before, both in Newtonian dynamics and relativistic dynamics.}


\textbf{A particle of mass m collides elastically at speed u with an identical particle at
rest. Show that in Newtonian dynamics the two particles after the collision always
emerge at right angles, whatever u or the scattering angle.}

Let $\vec p_1$ be the momentum of the probe particle, and $\vec p_2 = 0$ the momentum of the target particle, both before the collision. After the collision the probe and target particles are described by $\vec p_3$ and $\vec p_4$ respectively, then by conservation of 3-momentum we have $$\vec p_1 = \vec p_2 + \vec p_3 \Rightarrow \vec p_1^2 = \vec p_2^2 + \vec p_3^2 + 2 \vec p_2 \cdot \vec p_3.$$ Since kinetic energy and momentum are related by $E = \vec p^2 / 2m$, and we are told that the collision is elastic, then energy will be conserved according to $$\frac{\vec p_1^2}{2m_1} = \frac{\vec p_3^2}{2m_1} + \frac{\vec p_3^2}{2m_2}$$ but the particles are identical so $m_1 = m_2$ and $$\vec p_1^2 = \vec p_2^2 + \vec p_3^2$$ and so by substituting this in the above expression for conservation of momentum we have $$\vec p_2 \cdot \vec p_3 = 0$$ i.e. $\vec p_2$ and $\vec p_3$ are at right angles.

\textbf{Show that in relativistic dynamics, the angle $\alpha$ between the two particles after the collision is given by...}



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